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Suppose X and Y are two independent exponential variables. The mean of X is twice the mean of Y. If the probability of X exceeding 50 is 0.7788, what is the probability of Y exceeding 40

Sagot :

LammettHashidn

If X ~ Exponential(µ), then the mean of X is 1/µ. So if the mean of X is twice the mean of Y, then the mean of Y is 1/(2µ), so that Y ~ Exponential(2µ).

We're given that

P(X > 50) = 1 - P(X ≤ 50) = 1 - Fx (50) ≈ 0.7788

==>   Fx (50) = P(X ≤ 50) ≈ 0.2212

where Fx is the CDF of X, which is given for 0 ≤ x < ∞ to be

Fx (x) = 1 - exp(-µx)

Solve for µ :

1 - exp(-50µ) ≈ 0.2212   ==>   µ ≈ -ln(0.7788)/50 ≈ 0.005

Then we have

P (Y > 40) = 1 - P (Y ≤ 40) = 1 - Fy (40)

where Fy is the CDF of Y,

Fy (y) = 1 - exp(-2µy)

so that

P (Y > 40) ≈ 1 - exp(-2 × 0.005 × 40) ≈ 0.3297

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